#elementary operations

Definition of Elementary Operations

Elementary operations are operations that consist of the following:

1. Interchange the order in which the equations are listed. 2. Multiply any equation by a nonzero scalar. 3. Replace any equation with itself plus a multiple of another equation.

None of these elementary operations will change the set of solutions of the system of equations. Elementary operations are the key tool linear algebra uses to find solutions to systems of equations.

A linear system may involve many equations and many variables, where, for any size of system in any number of variables, the solution set is still the collection of solutions to the equations. The elementary operations do not change the set of solutions to the system of equations.

Theorem of Elementary Operations and Solutions The following theorem uses the notation Eᵢ to represent an equation, while bᵢ denotes a constant.

Suppose there is a system of two linear equations called 1.1, such as the following: E₁ = b₁ E₂ = b₂

Then the following systems, 1.2, 1.3 and 1.4, respectively, have the same solution as the system 1.1:

1.2. E₂ = b₂ E₁ = b₁

1.3. E₁ = b₁ kE₂ = kb₂ (for any scalar k, where k ≠ 0)

1.4 E₁ = b₁ E₂ + kE₁ = b₂ + kb₁ (for any scalar k)

The following proof shows how system 1.1 and system 1.2 have the same solution set:

Suppose that (x₁, …, xn) is a solution to the system 1.1.

It is clear that this solution set is the same solution set for system 1.2, because the system 1.2 is the same as system 1.1 but in different order, where changing the order does not effect the solution set, and so (x₁, …, xn) is a solution to system 1.2.

The following proof shows how system 1.1 and system 1.3 have the same solution set:

Suppose that (x₁, …, xn) is a solution to the system 1.1.

Notice that the only difference between the systems 1.1 and 1.3 is that the second equation involves multiplying the equation by the scalar k.

In algebra, multiplying the same number on both sides of an equation does not affect the solution, making system 1.3's solution equal to system 1.1's solution.

Additionally, we can multiply 1/k on both sides of the second equation in system 1.3, which leaves this system to be the exact same system as system 1.1.

The following proof shows how system 1.1 and system 1.4 have the same solution set:

Suppose that (x₁, …, xn) is a solution to the system 1.1.

Since (x₁, …, xn) solves the first equation in system 1.1, it solves the first equation in system 1.4, because it is the same equation.

Since (x₁, …, xn) solves the first and second equation in system 1.1, it solves both equations in system 1.3.

If E₂ was added with kE₁, then the following is true: E₂ + kE₁ = b₂ + kb₁

Therefore, if (x₁, …, xn) solves the first equation in system 1.4, then it solves the second equation in system 1.4.

Back Substitution The following is an example of a system of three equations in three variables using the process of back substitution:

Find the solution to the following system: x + 3y + 6z = 25 2x + 7y + 14z = 58 2y + 5z = 19

The above theorem claims that elementary operations will not change the solution set. In order to use back substitution, we make the first variable in the first equation have a coefficient of one using the second elementary operation, and then eliminate this variable in the second equation using the third elementary operation. Next, the new leading variable in the second equation needs a coefficient of one, so the second elementary operation is used. Then eliminate this variable in the third equation using the third elementary operation, leaving only one variable left in the third equation. Lastly, using the third equation, substitute into the second equation and then into the first equation.

1. We make the first variable in the first equation have a coefficient of one using the second elementary operation.

According to the first equation, x already has a coefficient of one.

2. Eliminate this variable in the second equation using the third elementary operation.

To eliminate this variable in the second equation, the second equation must be replaced by -2 times the first equation added to the second equation.

x + 3y + 6z = 25 2x + 7y + 14z = 58 2y + 5z = 19

x + 3y + 6z = 25 2x + 7y + 14z + (-2x - 6y - 12z) = 58 + (-50) 2y + 5z = 19

x + 3y + 6z = 25 y - 2z = 8 2y + 5z = 19

3. The new leading variable in the second equation needs a coefficient of one, so the second elementary operation is used.

According to the second equation, y already has a coefficient of one.

4. Eliminate this variable in the third equation using the third elementary operation, leaving only one variable left in the third equation.

To eliminate this variable in the third equation, the third equation must be replaced by -2 times the second equation added to the third equation.

x + 3y + 6z = 25 y - 2z = 8 2y + 5z = 19

x + 3y + 6z = 25 y - 2z = 8 2y + 5z + (-2y + 4z) = 19 + (-16)

x + 3y + 6z = 25 y - 2z = 8 z = 3

5. Using the third equation, substitute into the second equation and then into the first equation.

If z = 3: y - 2z = 8 y = 8 + 2(3) = 2

If z = 3, y = 2: x + 3y + 6z = 25 x = 25 - 3(2) - 6(3) = 25 - 6 - 18 = 1